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Math Help - Fields

  1. #1
    Forum Admin topsquark's Avatar
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    Fields

    Got another one for y'all. I got part of the way through this, but I'm stuck for a way to continue.

    Here's the question:

    "Let K be a field, and let R be a subring of K. Suppose that every element of K satisfies a polynomial whose coefficients are in R and whose highest degree coefficient is 1. Show that R is actually a subfield of K."

    The question goes on to give a couple of hints:
    1) Show that R is an integral domain
    2) Given a nonzero a \epsilon R we know that 1/a satisfies a suitable polynomial with coefficients in R. Manipulate this polynomial.

    Here's what I've got so far:

    Proof: R is an integral domain.
    Assume R is a subring of the field K. Then R is an integral domain because any element of R is also an element of K, thus the subring R is commutative and contains no zero-divisors. (End of proof.)

    For the second part, here's as far as I've gotten.

    By hypothesis, \forall a\epsilon K \exists \{r_i\} \subset R and n \epsilon Z_{+} such that:
    a^n+r_{n-1}a^{n-1}+...+r_0=0

    Now, since a \epsilon K and K is a field \exists 1/a \epsilon K. We may multiply the above equation by (1/a)^n:
    1+r_{n-1}(1/a)+...+r_0(1/a)^n=0
    or
    r_0(1/a)^n+r_1(1/a)^{n-1}+...+1=0.

    Again, since a \epsilon K and K is a field \exists 1/a \epsilon K, by hypothesis:
    (1/a)^m+q_{m-1}(1/a)^{m-1}+...+q_0=0; \{q_i\} \subset R
    where m does not need to be equal to n. We may multiply this equation by r_0:
    r_0(1/a)^m+r_0q_{m-1}(1/a)^{m-1}+...+r_0q_0=0.
    Compare this with
    r_0(1/a)^n+r_1(1/a)^{n-1}+...+1=0.

    I stated before that m was not necessarily equal to n, but these equations together show that the set of all possible values for m (given this a) contains the value n. So set m=n. Furthermore, R contains no zero-divisors, so the second equation is not trivial (that is, all of the coefficients are non-zero.) Finally, the leading coefficient is the same in both equations. Thus I conclude the polynomials are equal in the sense that ALL the coefficients of like terms are equal.

    In particular, comparison of the constant term shows r_0q_0=1 where both r_0,q_0 \epsilon R. Thus, by definition, q_0 is the multiplicative inverse of r_0.

    This is as far as I've gotten. I can show by this method that some elements of R have inverses in R, but not all of them. Only those r_0 values that come up in polynomials for given values of a will have inverses, not the whole set R.

    Any thoughts to finish this, or am I beating down a dead end?

    Thanks!
    -Dan
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by topsquark
    Got another one for y'all. I got part of the way through this, but I'm stuck for a way to continue.

    Here's the question:

    "Let K be a field, and let R be a subring of K. Suppose that every element of K satisfies a polynomial whose coefficients are in R and whose highest degree coefficient is 1. Show that R is actually a subfield of K."

    The question goes on to give a couple of hints:
    1) Show that R is an integral domain
    2) Given a nonzero a \epsilon R we know that 1/a satisfies a suitable polynomial with coefficients in R. Manipulate this polynomial.

    Here's what I've got so far:

    Proof: R is an integral domain.
    Assume R is a subring of the field K. Then R is an integral domain because any element of R is also an element of K, thus the subring R is commutative and contains no zero-divisors. (End of proof.)

    For the second part, here's as far as I've gotten.

    By hypothesis, \forall a\epsilon K \exists \{r_i\} \subset R and n \epsilon Z_{+} such that:
    a^n+r_{n-1}a^{n-1}+...+r_0=0

    Now, since a \epsilon K and K is a field \exists 1/a \epsilon K. We may multiply the above equation by (1/a)^n:
    1+r_{n-1}(1/a)+...+r_0(1/a)^n=0
    or
    r_0(1/a)^n+r_1(1/a)^{n-1}+...+1=0.

    Again, since a \epsilon K and K is a field \exists 1/a \epsilon K, by hypothesis:
    (1/a)^m+q_{m-1}(1/a)^{m-1}+...+q_0=0; \{q_i\} \subset R
    where m does not need to be equal to n. We may multiply this equation by r_0:
    r_0(1/a)^m+r_0q_{m-1}(1/a)^{m-1}+...+r_0q_0=0.
    Compare this with
    r_0(1/a)^n+r_1(1/a)^{n-1}+...+1=0.

    I stated before that m was not necessarily equal to n, but these equations together show that the set of all possible values for m (given this a) contains the value n. So set m=n. Furthermore, R contains no zero-divisors, so the second equation is not trivial (that is, all of the coefficients are non-zero.) Finally, the leading coefficient is the same in both equations. Thus I conclude the polynomials are equal in the sense that ALL the coefficients of like terms are equal.

    In particular, comparison of the constant term shows r_0q_0=1 where both r_0,q_0 \epsilon R. Thus, by definition, q_0 is the multiplicative inverse of r_0.

    This is as far as I've gotten. I can show by this method that some elements of R have inverses in R, but not all of them. Only those r_0 values that come up in polynomials for given values of a will have inverses, not the whole set R.

    Any thoughts to finish this, or am I beating down a dead end?

    Thanks!
    -Dan
    I was finally able to figure it out. I'll post it for the sake of completeness.

    Our story so far...

    The subring R of the field K is actually an integral domain, so R contains no zero-divisors. Additionally, the hypothesis assures us that \forall a \in K we have that \exists \{r_i \} \subset R and n \in \mathbb{Z}_{+} such that:
    a^n+r_{n-1}a^{n-1}+...+r_0=0.
    This leads to the result that r_0 has a multiplicative inverse in R.

    The problem at this point is that we need to prove that an inverse exists for all elements of R. I was finally able to get around that problem:

    Since a^n+r_{n-1}a^{n-1}+...+r_0=0, take any b \in R and form the product polynomial:
    (a+b)(a^n+r_{n-1}a^{n-1}+...+r_0)=0
    This produces a new polynomial:
    a^{n+1}+...+br_0=0.
    By the construction in the previous post (and as repeated above) the existence of this polynomial implies that the element br_0 has an inverse in R also. Recall that multiplication in R is commutative (since it is a subring of the field K.) Thus:
    \frac{1}{br_0} \in R \Rightarrow r_0 \frac{1}{br_0} \in R \Rightarrow \frac{1}{b} \in R \forall b \in R, since multiplication in R is closed by definition.

    So all elements of R have a multiplicative inverse, making the integral domain R a field.

    -Dan
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  3. #3
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    I think you can make it rather shorter. For r \in R, the inverse r^{-1} \in K satisfies a monic equation r^{-n} + a_{n-1}r^{-n+1} + \cdots + a_1 r^{-1} + a_0 = 0 with all the a_i \in R. Multiply by r^{n-1} to get r^{-1} = -a_{n-1} - \cdots - a_0 r^{n-1} and the RHS is in R since it is a ring.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by rgep
    I think you can make it rather shorter. For r \in R, the inverse r^{-1} \in K satisfies a monic equation r^{-n} + a_{n-1}r^{-n+1} + \cdots + a_1 r^{-1} + a_0 = 0 with all the a_i \in R. Multiply by r^{n-1} to get r^{-1} = -a_{n-1} - \cdots - a_0 r^{n-1} and the RHS is in R since it is a ring.
    Bugger.

    I hate it when I miss the easy way...

    Thanks for the tip!

    -Dan
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